1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Concavity of trig f'n

  1. Apr 13, 2013 #1
    1. The problem statement, all variables and given/known data
    f(x) = cos^2(x) - 2sin(x)

    0≤x≤2∏


    2. Relevant equations



    3. The attempt at a solution
    f'(x)=-2[(cosX)(sinX) + cosX]

    f''(x) = -2[ (cos^2(x)-sin^2(x)-sinX]

    I know there's an identity there for cos2x but it doesn't seem to help me. I also tried to go the other way and use the pythagorean identity and got 2sin^2(x)-sinX=1

    but none of these make it easy to tell where f''(x)=0

    Further, the book says something like if f''(c)=0 you get no information about concavity.
    What do they mean? Can't you simply change your c to another number until you get something that's greater or less than 0?
     
  2. jcsd
  3. Apr 13, 2013 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What are you trying to determine? State the problem you're working on ... precisely.

    From the title you gave, it looks as if you're trying to determine the concavity of f '(x) .
     
  4. Apr 13, 2013 #3
    Intervals of concavity and the points of inflection of f(x).
     
  5. Apr 13, 2013 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    not even a sentence


    I can guess what you're being asked. Your reply was far from precise.


    You can write the first derivative as
    [itex]\displaystyle \
    f'(x)=-\sin(2x) -2\cos(x)\ .[/itex]​

    That makes the second derivative a bit easier to find.
     
  6. Apr 14, 2013 #5
    That's all the book says: find the intervals of concavity and the points of inflection and then it lists a bunch of functions.
    That's not even my problem - I've already taken the derivatives. My problem is I can't figure out when f''(x)=0
     
  7. Apr 14, 2013 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do you see how this reply of yours is a big help for anybody trying to help you? Now we know what you are trying to find and what you were having trouble with.

    If you write the first derivative in the form I suggested, ##\ \displaystyle \
    f'(x)=-\sin(2x) -2\cos(x)\,, \ ## then the second derivative is
    ## \displaystyle
    f''(x)=-2\cos(2x)+2\sin(x)\ .
    ##​

    Now use the identity for cos(2x) which involves only the sine.
    ##\displaystyle
    \cos(2x)=1-2\sin^2(x)##​

    Setting ##\ f''(x)\ ## equal to zero then gives you an equation that's quadratic in sin(x). Solve that like you would for a trig class.
     
  8. Apr 15, 2013 #7

    Mark44

    Staff: Mentor

    Your first post in this thread should have been something like this:
    1. The problem statement, all variables and given/known data

    find the intervals of concavity and the points of inflection of f(x) = cos^2(x) - 2sin(x) on [0, 2##\pi##].

    Sammy was trying to get you to provide a precise statement of your problem. If the problem is not stated well, it can cause people who are trying to help you to flounder around trying to understand what the problem is.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Concavity of trig f'n
Loading...